Related papers: Super congruences and Euler numbers

The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2x^k\ \ (n=0,1,2,\ldots).$$ (Those $A_n=A_n(1)$ are Ap\'ery numbers.) Let $p$ be an odd prime. We show that…

In this paper, we mainly prove a congruence conjecture of Z.-W. Sun \cite{Sjnt}: Let $p>5$ be a prime. Then $$ \sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k^2}{k16^k}\equiv-\frac{21}2H_{p-1}\pmod{p^4}, $$ where $H_n$ denotes the $n$-th harmonic…

In this paper we prove the supercongruence $$\sum_{n=0}^{(p-1)/2}\frac{6n+1}{256^n}\binom{2n}n^3\equiv p(-1)^{(p-1)/2}+(-1)^{(p-1)/2}\frac{7}{24}p^4B_{p-3}\pmod{p^5}$$ for any prime $p>3$, which was conjectured by Sun in 2019.

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence, which is $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n$-th Bernoulli number. In this paper, we will…

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper we solve some conjectures of Z.W. Sun concerning $\sum_{k=0}^{p-1}\binom{2k}k^3/m^k\pmod{p^2}$, $\sum_{k=0}^{p-1}\binom{2k}k\b{4k}{2k}/m^k\pmod p$ and…

The Ap\'ery numbers $A_n$ and the Franel numbers $f_n$ are defined by $$A_n=\sum_{k=0}^{n}{\binom{n+k}{2k}}^2{\binom{2k}{k}}^2\ \ \ \ \ {\rm and }\ \ \ \ \ \ f_n=\sum_{k=0}^{n}{\binom{n}{k}}^3(n=0, 1, \cdots,).$$ In this paper, we prove…

Let $p>3$ be a prime, and let $a$ be a rational $p$-adic integer, using WZ method we establish the congruences modulo $p^3$ for $$\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k},$$ where $$w(k)=1,\frac 1{k+1},\frac…

The Franel numbers given by $f_n=\sum_{k=0}^n\binom{n}{k}^3$ ($n=0,1,2,\ldots$) play important roles in both combinatorics and number theory. In this paper we initiate the systematic investigation of fundamental congruences for the Franel…

Let $p$ be an odd prime, and let $a$ be a rational $p$-adic integer with $a\not\equiv 0\pmod p$. In this paper, using WZ method we establish the congruences for $\sum_{k=0}^{p-1} \binom ak^2(-1)^k(1-\frac 2ak)$ modulo $p^2$ and…

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

Let $n$ be a nonnegative integer. The $n$-th Ap\'{e}ry number is defined by $$ A_n:=\sum_{k=0}^n\binom{n+k}{k}^2\binom{n}{k}^2. $$ Z.-W. Sun ever investigated the congruence properties of Ap\'{e}ry numbers and posed some conjectures. For…

For various positive integers $k$, the sums of $k$th powers of the first $n$ positive integers, $S_k(n+1)=1^k+2^k+...+n^k$, have got to be some of the most popular sums in all of mathematics. In this note we prove that for each $k\ge 2 $$…

The Ap\'ery numbers $A_n$ and central Delannoy numbers $D_n$ are defined by $$A_n=\sum_{k=0}^{n}{n+k\choose 2k}^2{2k\choose k}^2, \quad D_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}. $$ Motivated by some recent work of Z.-W. Sun, we prove…

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. Z. W. Sun conjectured that for any prime $p\ge 5$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k\cdot 2^k} \equiv7/24pB_{p-3}\pmod{p^2}. $$ This conjecture is recently…

We evaluate some series with summands involving a single binomial coefficient $\binom{6k}{3k}$. For example, we prove that $$\sum_{k=0}^\infty\frac{(63k^2+78k+22)8^k}{(2k+1)(6k+1)(6k+5)\binom{6k}{3k}}=\frac{3\pi}2.$$ Motivated by Galois…

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive…

We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that $$ \sum_{k=0}^{[n/4]}(-1)^k {n\choose 4k}\frac{B_{n-4k}(z) }{2^{6k}} =\frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^k…

Let p be any odd prime. We mainly show that $$\sum_{k=1}^{p-1}binomial(3k,k)*2^k/k=0 (mod p)$$ and $$\sum_{k=1}^{p-1}2^{k-1}C_k^{(2)}=(-1)^{(p-1)/2}-1 (mod p),$$ where $C_k^{(2)}=binomial(3k,k)/(2k+1)$ is the $k$th Catalan number of order…

Let $M_n$ and $T_n$ denote the $n$th Motzkin number and the $n$th central trinomial coefficient respectively. We prove that for any prime $p\ge 5$, \begin{align*} &\sum_{k=0}^{p-1}M_k^2\equiv…

Let $p$ be an odd prime. It is well known that $F_{p-(\frac p5)}\equiv 0\pmod{p}$, where $\{F_n\}_{n\ge0}$ is the Fibonacci sequence and $(-)$ is the Jacobi symbol. In this paper we show that if $p\not=5$ then we may determine $F_{p-(\frac…